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2328. Number of Increasing Paths in a Grid 👍

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int countPaths(vector<vector<int>>& grid) {
    m = grid.size();
    n = grid[0].size();
    int ans = 0;
    // dp[i][j] := # of increasing paths starting from (i, j)
    dp.resize(m, vector<int>(n, -1));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        ans += dfs(grid, i, j);
        ans %= kMod;
      }

    return ans;
  }

 private:
  static constexpr int kMod = 1'000'000'007;
  const vector<int> dirs{0, 1, 0, -1, 0};
  int m;
  int n;
  vector<vector<int>> dp;

  int dfs(const vector<vector<int>>& grid, int i, int j) {
    if (dp[i][j] != -1)
      return dp[i][j];

    dp[i][j] = 1;  // Current cell contributes 1 length

    for (int k = 0; k < 4; ++k) {
      const int x = i + dirs[k];
      const int y = j + dirs[k + 1];
      if (x < 0 || x == m || y < 0 || y == n)
        continue;
      if (grid[x][y] <= grid[i][j])
        continue;
      dp[i][j] += dfs(grid, x, y);
      dp[i][j] %= kMod;
    }

    return dp[i][j];
  }
};

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class Solution {
  public int countPaths(int[][] grid) {
    m = grid.length;
    n = grid[0].length;
    int ans = 0;
    // dp[i][j] := # of increasing paths starting from (i, j)
    dp = new int[m][n];
    Arrays.stream(dp).forEach(A -> Arrays.fill(A, -1));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j) {
        ans += dfs(grid, i, j);
        ans %= kMod;
      }

    return ans;
  }

  private static final int kMod = 1_000_000_007;
  private final int[] dirs = {0, 1, 0, -1, 0};
  private int m;
  private int n;
  private int[][] dp;

  private int dfs(int[][] grid, int i, int j) {
    if (dp[i][j] != -1)
      return dp[i][j];

    dp[i][j] = 1; // Current cell contributes 1 length

    for (int k = 0; k < 4; ++k) {
      final int x = i + dirs[k];
      final int y = j + dirs[k + 1];
      if (x < 0 || x == m || y < 0 || y == n)
        continue;
      if (grid[x][y] <= grid[i][j])
        continue;
      dp[i][j] += dfs(grid, x, y);
      dp[i][j] %= kMod;
    }

    return dp[i][j];
  }
}

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class Solution:
  def countPaths(self, grid: List[List[int]]) -> int:
    kMod = 1_000_000_007
    m = len(grid)
    n = len(grid[0])
    dirs = [0, 1, 0, -1, 0]

    # dp(i, j) := # of increasing paths starting from (i, j)
    @functools.lru_cache(None)
    def dp(i: int, j: int) -> int:
      ans = 1  # Current cell contributes 1 length
      for k in range(4):
        x = i + dirs[k]
        y = j + dirs[k + 1]
        if x < 0 or x == m or y < 0 or y == n:
          continue
        if grid[x][y] <= grid[i][j]:
          continue
        ans += dp(x, y)
        ans %= kMod
      return ans

    return sum(dp(i, j)
               for i in range(m)
               for j in range(n)) % kMod